1. Given the series Sn = 1/3 + 1/15 + 1/35 + 1/63 + ......+1/(4n^2-1),
a) calculate S1, S2, S3, and S4.
b) Derive a formula for Sn.
c) Prove that M.I. that the formula for Sn is correct for all +ve integers n.
2. If x and y are itnegers, prove that x^(2n-1) + y^(2n-1) is divisible by x+y for all 'n'.
Challenging Maths
2007-07-13 11:12 pm
回答 (2)
2007-07-13 11:44 pm
✔ 最佳答案
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參考: My Maths knowledge
2007-07-13 11:31 pm
a)
S1=1/3
S2=2/5
S3=3/7
S4=4/9
b)
Sn=1/(1x3)+1/(3x5)+1/(5x7)+....+1/(2n-1)(2n+1)
= (1/2) x (1-1/(2n+1))
= n/2n+1
c)
if n = 1
S1=1/3= n/2n+1
if n = k
Sk is assumed to be right
if n = k+1
1/(1x3)+1/(3x5)+1/(5x7)+....+1/(2k-1)(2k+1) + 1/(2k+1)(2k+3)
=k/(2k+1)+1 / (2k+1)(2k+3)
=k(2k+3)+1 / (2k+1)(2k+3)
=(2k+1)(k+1) / (2k+1)(2k+3)
=(k+1) / (2k+3)
= n/2n+1
by the principle of tmathematical induction,S(n) is true for all integers n
2) n is an integer
(2n-1) is not divisible by 2
x^(2n-1) + y^(2n-1)
= (x+y)(x^(2n-2)-x^(2n-3)y+x^(2n-4)y^2-...........+y^(2n-2))
.'. if x & y are integer , x^(2n-1) + y^(2n-1) is divisible by x+y for all 'n'
"itnegers" should be integer
S1=1/3
S2=2/5
S3=3/7
S4=4/9
b)
Sn=1/(1x3)+1/(3x5)+1/(5x7)+....+1/(2n-1)(2n+1)
= (1/2) x (1-1/(2n+1))
= n/2n+1
c)
if n = 1
S1=1/3= n/2n+1
if n = k
Sk is assumed to be right
if n = k+1
1/(1x3)+1/(3x5)+1/(5x7)+....+1/(2k-1)(2k+1) + 1/(2k+1)(2k+3)
=k/(2k+1)+1 / (2k+1)(2k+3)
=k(2k+3)+1 / (2k+1)(2k+3)
=(2k+1)(k+1) / (2k+1)(2k+3)
=(k+1) / (2k+3)
= n/2n+1
by the principle of tmathematical induction,S(n) is true for all integers n
2) n is an integer
(2n-1) is not divisible by 2
x^(2n-1) + y^(2n-1)
= (x+y)(x^(2n-2)-x^(2n-3)y+x^(2n-4)y^2-...........+y^(2n-2))
.'. if x & y are integer , x^(2n-1) + y^(2n-1) is divisible by x+y for all 'n'
"itnegers" should be integer
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